Plotting a System of Two Second-Order Differential Equations
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I'm trying to reduce a system of two second-order differential equations into a system of first-order equations, solve them, and plot the result. The differential equations are x1'' = -(k1*x1 - k2*(x1 - x2))/m and x2'' = -(k2(x2 - x1))/m2.
This is what I have so far:
[t, y] = ode45(@func,[0 100],[3, 0])
figure (9)
plot(t,y(:,1),t,y(:,3))
function dy = func(t,y)
m1 = 2;
m2 = 4;
k1 = 3;
k2 = 5;
dy = zeros(4,1)
dy(1) = y(2);
dy(2) = -(-(k1 + k2)*y(1)+k2*y(3))./m1;
dy(3) = y(4);
dy(4) = (k1*y(1) - k2*y(3))./m2;
end
It says that there is an error in line 1 and errors in ode45 and odearguments. There is also an error in line 11 and that "index exceeds the number of array elements."
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James Tursa
2019-12-19
编辑：James Tursa
2019-12-19
You've got a 4th order system, so your initial state must contain four elements including the x1' and x2', not two. E.g.,
[t, y] = ode45(@func,[0 100],[3, 0, 0, 0]); % x1, x1', x2, x2' in initial conditions
Also, your dy(2) and dy(4) don't match your posted derivative equations, so you should double check and correct those.
2 个评论
Matlab12345
2019-12-19
James Tursa
2019-12-19
编辑：James Tursa
2019-12-19
Based on what you wrote, this equation
x1'' = -(k1*x1 - k2*(x1 - x2))/m
translates to this code
dy(2) = -(k1*y(1) - k2*(y(1) - y(3))) / m1;
and this equation
x2'' = -(k2(x2 - x1))/m2
translates to this code
dy(4) = -k2*(y(3) - y(1)) / m2;
I don't know the point of rearranging the terms in your code since it just makes it more difficult to compare directly to the derivative equation. And in fact it looks like you still don't have the dy(2) code correct, so you might consider just doing the straghtforward translation and using my supplied code.
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